2 Elementary probability

2

Elementary probability

This first chapter devoted to probability theory contains the basic definitions and con-cepts in this field, without the formalism of measure theory. However, the range ofproblems that can be solved by using the formulas of elementary probability is verybroad, particularly in combinatorial analysis. Therefore, it is necessary to do numerousexercises in order to master these basic concepts.

2.1 Random experiments

A random experiment is an experiment that, at least theoretically, may be repeated asoften as we want and whose outcome cannot be predicted, for example, the roll of adie. Each time the experiment is repeated, an elementary outcome is obtained. The setof all elementary outcomes of a random experiment is called the sample space, which isdenoted by Ω.

Sample spaces may be discrete or continuous.(a) Discrete sample spaces. (i) Firstly, if the number of possible outcomes is finite.For example, if a die is rolled and the number that shows up is noted, then Ω =1, 2, . . . , 6.

ii) Secondly, if the number of possible outcomes is countably infinite, which meansthat there is an infinite number of possible outcomes, but these outcomes can be putin a one-to-one correspondence with the positive integers. For example, if a die is rolleduntil a “6” is obtained, and the number of rolls made before getting this first “6” iscounted, then we have that Ω = 0, 1, 2, . . .. This set is equivalent to the set of allnatural integers 1, 2, . . ., because we can associate the natural number k+1 with eachelement k = 0, 1, . . . of Ω.(b) Continuous sample spaces. If the sample space contains one or many intervals,the sample space is then uncountably infinite. For example, a die is rolled until a “6”is obtained and the time needed to get this first “6” is recorded. In this case, we havethat Ω = t ∈ R : t > 0 [or Ω = (0,∞)].

© Springer Science + Business Media, LLC 2009and Technology, DOI: 10.1007/978-0-387-74995-2_2,

27M. Lefebvre, Basic Probability Theory with Applications, Springer Undergraduate Texts in Mathematics

28 2 Elementary probability

2.2 Events

Definition 2.2.1. An event is a set of elementary outcomes. That is, it is a subset ofthe sample space Ω. In particular, every elementary outcome is an event, and so is thesample space itself.

Remarks. (i) An elementary outcome is sometimes called a simple event, whereas acompound event is made up of at least two elementary outcomes.(ii) To be precise, we should distinguish between the elementary outcome ω, which isan element of Ω, and the elementary event ω ⊂ Ω.(iii) The events are denoted by A, B, C, and so on.

Definition 2.2.2. Two events, A and B, are said to be incompatible (or mutuallyexclusive) if their intersection is empty. We then write that A ∩ B = ∅.

Example 2.2.1. Consider the experiment that consists in rolling a die and recordingthe number that shows up. We have that Ω = 1, 2, 3, 4, 5, 6. We define the events

A = 1, 2, 4, B = 2, 4, 6 and C = 3, 5.

We have:A ∪B = 1, 2, 4, 6, A ∩B = 2, 4 and A ∩ C = ∅.

Therefore, A and C are incompatible events. Moreover, we may write that A′ = 3, 5, 6,where the symbol ′ denotes the complement of the event.

To represent a sample space and some events, we often use a Venn diagram as inFigure 2.1. In general, for three events we have the diagram in Figure 2.2.

A B C

1 24

6 3 5

Ω

Fig. 2.1. Venn diagram for Example 2.2.1.

2.3 Probability 29

A B

C

ω

1

2

3 4

5 6

78

ω = A B C1

ω = A B C'

ω = A B' C

ω = A' B C

ω = A B' C'

ω = A' B C'

ω = A' B' C

ω = A' B' C'

2

3

4

5

6

7

8

U U

U U

U U

U U

UU

UU

UU

UU

Ω

ω

ω ω

ω ωω

ω

Fig. 2.2. Venn diagram for three arbitrary events.

2.3 Probability

Definition 2.3.1. The probability of an event A ⊂ Ω, denoted by P [A], is a realnumber obtained by applying to A the function P that possesses the following properties:

(i) 0 ≤ P [A] ≤ 1;(ii) if A = Ω, then P [A] = 1;(iii) if A = A1 ∪A2 ∪ · · · ∪An, where A1, . . . , An are incompatible events, then we maywrite that

P [A] =n∑

i=1

P [Ai] for n = 2, 3, . . . ,∞.

Remarks. (i) Actually, we only have to write that P [A] ≥ 0 in the definition, becausewe can show that

P [A] + P [A′] = 1,

which implies that P [A] = 1− P [A′] ≤ 1.(ii) We also have the following results:

P [∅] = 0 and P [A] ≤ P [B] if A ⊂ B.

(iii) The definition of the probability of an event is motivated by the notion of relativefrequency. For example, suppose that the random experiment that consists in rolling adie is repeated a very large number of times, and that we wish to obtain the probabilityof any of the possible outcomes of this experiment, namely, the integers 1, 2, . . . , 6. Therelative frequency of the elementary event k is the quantity fk(n) defined by

fk(n) =Nk(n)

n,


where Nk(n) is the number of times that the possible outcome k occurred among then rolls of the die. We can write that

0 ≤ fk(n) ≤ 1 for k = 1, 2, . . . , 6

and that6∑

k=1

fk(n) = 1.

Indeed, we obviously have that Nk(n) ∈ 0, 1, . . . , n, so that fk(n) belongs to [0, 1],and

6∑k=1

fk(n) =N1(n) + · · ·+ N6(n)

n=

n

n= 1.

Furthermore, if A is an event containing two possible outcomes, for instance “1” and“2,” then

fA(n) = f1(n) + f2(n),

because the outcomes 1 and 2 cannot occur on the same roll of the die.Finally, the probability of the elementary event k can theoretically be obtained

by taking the limit of fk(n) as the number n of rolls tends to infinity:

P [k] = limn→∞

fk(n).

The probability of an arbitrary event can be expressed in terms of the relative frequencyof this event, thus it is logical that the properties of probabilities more or less mimicthose of relative frequencies.

Sometimes, the probability of an elementary outcome is simply equal to 1 dividedby the total number of elementary outcomes. In this case, the elementary outcomes aresaid to be equiprobable (or equally likely). For example, if a fair (or unbiased) die isrolled, then we have that P [1] = P [2] = · · · = P [6] = 1/6.

If the elementary outcomes ri are not equiprobable, we can (try to) make use of thefollowing formula:

P [A] =∑ri∈A

P [ri].

However, this formula is only useful if we know the probability of all the elementaryoutcomes ri that constitute the event A.

Now, if A and B are incompatible events, then we deduce from the third propertyof P [ · ] that P [A ∪ B] = P [A] + P [B]. If A and B are not incompatible, we can show(see Figure 2.3) that

P [A ∪B] = P [A] + P [B]− P [A ∩B].

2.3 Probability 31

A B

A

U

B

Ω

P[A U B] = P[A] + P[B] - P[A B]

U

Fig. 2.3. Probability of the union of two arbitrary events.

Similarly, in the case of three arbitrary events, we have:

P [A∪B ∪C]=P [A] + P [B] + P [C]−P [A∩B]−P [A∩C]−P [B ∩C] + P [A∩B ∩C].

Example 2.3.1. The three most popular options for a certain model of new car are A:automatic transmission, B: V6 engine, and C: air conditioning. Based on the previoussales data, we may suppose that P [A] = 0.70, P [B] = 0.75, P [C] = 0.80, P [A ∪ B]= 0.80, P [A ∪ C] = 0.85, P [B ∪ C] = 0.90, and P [A ∪ B ∪ C] = 0.95, where P [A]denotes the probability that an arbitrary buyer chooses option A, and so on. Calculatethe probability of each of the following events:(a) the buyer chooses at least one of the three options;(b) the buyer does not choose any of the three options;(c) the buyer chooses only air conditioning;(d) the buyer chooses exactly one of the three options.

Solution. (a) We seek P [A ∪B ∪ C] = 0.95 (by assumption).(b) We now seek P [A′ ∩B′ ∩ C ′] = 1− P [A ∪B ∪ C] = 1− 0.95 = 0.05.(c) The event whose probability is requested is A′ ∩B′ ∩ C. We can write that

P [A′ ∩B′ ∩ C] = P [A ∪B ∪ C]− P [A ∪B] = 0.95− 0.8 = 0.15.

(d) Finally, we want to calculate

P [(A ∩B′ ∩ C ′) ∪ (A′ ∩B ∩ C ′) ∪ (A′ ∩B′ ∩ C)]inc.= P [A ∩B′ ∩ C ′] + P [A′ ∩B ∩ C ′] + P [A′ ∩B′ ∩ C]= 3P [A ∪B ∪ C]− P [A ∪B]− P [A ∪ C]− P [B ∪ C]= 3(0.95)− 0.8− 0.85− 0.9 = 0.3.

Remarks. (i) The indication “inc.” above the “=” sign means that the equality is truebecause of the incompatibility of the events. We use this type of notation often in thisbook to justify the passage from an expression to another.


(ii) The probability of each of the eight elementary outcomes is indicated in the diagramof Figure 2.4. First, we calculate

P [A ∩B] = P [A] + P [B]− P [A ∪B] = 0.7 + 0.75− 0.8 = 0.65.

Likewise, we have:

P [A ∩ C] = 0.7 + 0.8− 0.85 = 0.65,

P [B ∩ C] = 0.75 + 0.8− 0.9 = 0.65,

P [A ∩B ∩ C] = P [A ∪B ∪ C]− P [A]− P [B]− P [C]+P [A ∩B] + P [A ∩ C] + P [B ∩ C]

= 0.95− 0.7− 0.75− 0.8 + 3(0.65) = 0.65.

A B

C

Ω

0.05 0

0.15

00

0.10

0.65

0.05

Fig. 2.4. Venn diagram for Example 2.3.1.

2.4 Conditional probability

Definition 2.4.1. The conditional probability of event A, given that event B oc-curred, is defined (and denoted) by (see Figure 2.5)

P [A | B] =P [A ∩B]

P [B]if P [B] > 0. (2.1)

From the above definition, we obtain the multiplication rule:

P [A ∩B] = P [A | B]P [B] if P [B] > 0 (2.2)

andP [A ∩B] = P [B | A]P [A] if P [A] > 0.

2.4 Conditional probability 33

A B

A

U

B

Ω

B

A

U

B

Fig. 2.5. Notion of conditional probability.

Definition 2.4.2. Let A and B be two events such that P [A]P [B] > 0. We say that Aand B are independent events if

P [A | B] = P [A] or P [B | A] = P [B]. (2.3)

We deduce from the multiplication rule that A and B are independent if and onlyif (iff)

P [A ∩B] = P [A]P [B]. (2.4)

Actually, this equation is the definition of independence of events A and B in thegeneral case when we can have that P [A]P [B] = 0. However, Definition 2.4.2 is moreintuitive, whereas the general definition of independence given by Formula (2.4) is purelymathematical.

In general, the events A1, A2, . . . , An are independent iff

P [Ai1 ∩ · · · ∩Aik] =

k∏j=1

P [Aij]

for k = 2, 3, . . . , n, where Ail6= Aim if l 6= m.

Remark. If A and B are two incompatible events, then they cannot be independent,unless P [A]P [B] = 0. Indeed, in the case when P [A]P [B] > 0, we have:

P [A | B] =P [A ∩B]

P [B]=

P [∅]P [B]

=0

P [B]= 0 6= P [A].

Example 2.4.1. A device is constituted of two components, A and B, subject to failures.The components are connected in parallel (see Figure 2.6) and are not independent. Weestimate the probability of a failure of component A to be 0.2 and that of a failure ofcomponent B to be 0.8 if component A is down, and to 0.4 if component A is not down.

(a) Calculate the probability of a failure (i) of component B and (ii) of the device.


A

B

Fig. 2.6. System for part (a) of Example 2.4.1.

Solution. Let A (resp., B) be the event “component A (resp., B) is down.” By assump-tion, we have that P [A] = 0.2, P [B | A] = 0.8, and P [B | A′] = 0.4.

(i) We may write (see Figure 2.7) that

A B

A

U

B

Ω

U

A' B

Fig. 2.7. Venn diagram for part (a) of Example 2.4.1.

P [B] = P [A ∩B] + P [A′ ∩B] = P [B | A]P [A] + P [B | A′]P [A′]= (0.8)(0.2) + (0.4)(0.8) = 0.48.

(ii) We seek P [Device failure] = P [A ∩B] = P [B | A]P [A] = 0.16.

(b) In order to increase the reliability of the device, a third component, C, is addedto the system in such a way that components A, B, and C are connected in parallel(see Figure 2.8). The probability that component C fails is equal to 0.2, independentlyfrom the state (up or down) of components A and B. Calculate the probability that thedevice made up of components A, B, and C breaks down.

Solution. By assumption, P [C] = 0.2 and C is independent of A and B. Let F be theevent “the subsystem made up of components A and B fails.” We can write that

P [F ∩ C] ind.= P [A ∩B]P [C](a) (ii)

= (0.16)(0.2) = 0.032.

2.5 Total probability 35

A

B

C

Fig. 2.8. System for part (b) of Example 2.4.1.

2.5 Total probability

Let B1, B2, . . . , Bn be incompatible and exhaustive events; that is, we have:

Bi ∩Bj = ∅ if i 6= j andn⋃

i=1

Bi = Ω.

We say that the events Bi constitute a partition of the sample space Ω. It follows that

P

[n⋃

i=1

Bi

]=

n∑i=1

P [Bi] = P [Ω] = 1.

Now, let A be an arbitrary event. We can write that (see Figure 2.9)

P [A] =n∑

i=1

P [A ∩Bi] =n∑

i=1

P [A | Bi]P [Bi] (2.5)

(the second equality above being valid when P [Bi] > 0, for i = 1, 2, . . . , n).Remark. This formula is sometimes called the law of total probability.

Finally, suppose that we wish to calculate P [Bi | A], for i = 1, . . . , n. We have:

P [Bi | A] =P [Bi ∩A]

P [A]=

P [A | Bi]P [Bi]∑nj=1 P [A ∩Bj ]

=P [A | Bi]P [Bi]∑n

j=1 P [A | Bj ]P [Bj ]. (2.6)

This formula is called Bayes’ formula.

Remark. We also have (Bayes’ rule):

P [B | A] =P [A | B]P [B]

P [A]if P [A]P [B] > 0. (2.7)


B

BBA

1 2

3

AA

A

BB

B

UU

U

12

3

U

P[A]=P[A B ]+P[A B ]+P[A B ]

Ω

U U

21 3

Fig. 2.9. Example of the law of total probability with n = 3.

Example 2.5.1. Suppose that machines M1, M2, and M3 produce, respectively, 500,1000, and 1500 parts per day, of which 5%, 6%, and 7% are defective. A part producedby one of these machines is taken at random, at the end of a given workday, and it isfound to be defective. What is the probability that it was produced by machine M3?

Solution. Let Ai be the event “the part taken at random was produced by machineMi,” for i = 1, 2, 3, and let D be “the part taken at random is defective.” We seek

P [A3 | D] =P [D | A3]P [A3]∑3i=1 P [D | Ai]P [Ai]

=(0.07)

(15003000

)(0.05)

(16

)+ (0.06)

(13

)+ (0.07)

(12

)=

105190

' 0.5526.

2.6 Combinatorial analysis

Suppose that we perform a random experiment that can be divided into two steps. Onthe first step, outcome A1 or outcome A2 may occur. On the second step, either ofoutcomes B1, B2, or B3 may occur. We can use a tree diagram to describe the samplespace of this random experiment, as in Figure 2.10.

Example 2.6.1. Tests conducted with a new breath alcohol analyzer enabled us toestablish that (i) 5 times out of 100 the test proved positive even though the personsubjected to the test was not intoxicated; (ii) 90 times out of 100 the test provedpositive and the person tested was really intoxicated. Moreover, we estimate that 1% ofthe persons subjected to the test are really intoxicated. Calculate the probability that(a) the test will be positive for the next person subjected to it;(b) a given person is intoxicated, given that the test is positive.

2.6 Combinatorial analysis 37

1st step

B

2nd step

A

AB

B

B

BB

1

3

2

3

2

1

1

2

Fig. 2.10. Example of a tree diagram.

Solution. Let A be the event “the test is positive” and let E be “the person subjected tothe test is intoxicated.” From the above assumptions, we can construct the tree diagramin Figure 2.11, where the marginal probabilities of events E and E′ are written abovethe branches, as well as the conditional probabilities of events A and A′, given that Eor E′ occurred. Furthermore, we know by the multiplication rule that the product ofthese probabilities is equal to the probability of the intersections E ∩A, E ∩A′, and soon.

E

E'A

A'

A

A'0.95

0.01

0.05

0.10

0.90

0.99

P[E A] = 0.009

P[E A'] = 0.001

P[E' A] = 0.0495

P[E' A'] = 0.9405

U

U

U

U

Fig. 2.11. Tree diagram in Example 2.6.1.

(a) We have:P [A] = P [E ∩A] + P [E′ ∩A] = 0.0585.

(b) We calculate

P [E | A] =P [E ∩A]

P [A](a)=

0.0090.0585

' 0.1538.

Note that this probability is very low. If we assume that 60% of the individuals sub-jected to the test are intoxicated (rather than 1%), then we find that P [A] becomes0.56 and P [E | A] ' 0.9643, which is much more reasonable. Therefore, this breath


alcohol analyzer is only efficient if we use it for individuals who are suspected of beingintoxicated.

Remark. In general, if a random experiment comprises k steps and if there are nj possibleoutcomes on the jth step, for j = 1, . . . , k, then there are n1 × · · · × nk elementaryoutcomes in the sample space. This is known as the multiplication principle.

Suppose now that we have n distinct objects and that we take, at random andwithout replacement, r objects among them, where r ∈ (0, )1, . . . , n. The number ofpossible arrangements is given by

n× (n− 1)× · · · × [n− (r − 1)] =n!

(n− r)!:= Pn

r . (2.8)

The symbol Pnr designates the number of permutations of n distinct objects taken r at

a time. The order of the objects is important.

Remarks. (i) Reminder. We have that n! = 1×2×· · ·×n, for n = 1, 2, 3, . . . , and 0! = 1,by definition.(ii) Taking r objects without replacement means that the objects are taken one at atime and that a given object cannot be chosen more than once. This is equivalent totaking the r objects all at once. In the case of sampling with replacement, any objectcan be chosen up to r times.

Example 2.6.2. If we have four different letters (for instance, a, b, c, and d), then wecan form

P 43 =

4!(4− 3)!

=4!1!

= 24

different three-letter “words” if each letter is used at most once. We can use a treediagram to draw the list of words.

Finally, if the order of the objects is not important, then the number of ways to take,at random and without replacement, r objects among n distinct objects is given by

n× (n− 1)× · · · × [n− (r − 1)]r!

=n!

r!(n− r)!:= Cn

r ≡(

n

r

)(2.9)

for r ∈ (0, )1, . . . , n. The symbol Cnr , or

(nr

), designates the number of combinations

of n distinct objects taken r at a time.

Remarks. (i) Each combination of r objects enables us to form r! different permutations,because

P rr =

r!(r − r)!

=r!0!

= r!.

(ii) Moreover, it is easy to check that Cnr = Cn

n−r.

2.7 Exercises for Chapter 2 39

Example 2.6.3. Three parts are taken, at random and without replacement, among 10parts, of which 2 are defective. What is the probability that at least 1 defective part isobtained?

Solution. Let F be the event “at least one part is defective among the three partstaken at random.” We can write that

P [F ] = 1− P [F ′] = 1− C20 · C8

3

C103

= 1−1 · 8!

3!5!10!3!7!

= 1− 6× 79× 10

=815

= 0.53.

2.7 Exercises for Chapter 2

Solved exercises

Question no. 1We consider the following random experiment: a fair die is rolled; if (and only if)

a “6” is obtained, the die is rolled a second time. How many elementary outcomes arethere in the sample space Ω?

Question no. 2Let Ω = e1, e2, e3, where P [ei] > 0, for i = 1, 2, 3. How many different partitions

of Ω, excluding the partition ∅, Ω can be formed?

Question no. 3A fair die is rolled twice, independently. Knowing that an even number was obtained

on the first roll, what is the probability that the sum of the two numbers obtained isequal to 4?

Question no. 4Suppose that P [A] = P [B] = 1/4 and that P [A | B] = P [B]. Calculate P[A ∩B′].

Question no. 5A system is made up of three independent components. It operates if at least two

of the three components operate. If the reliability of each component is equal to 0.95,what is the reliability of the system?

Question no. 6Suppose that P [A ∩B] = 1/4, P [A | B′] = 1/8, and P [B] = 1/2. Calculate P [A].

Question no. 7Knowing that we obtained at least once the outcome “heads” in three independent

tosses of a fair coin, what is the probability that we obtained “heads” three times?


Question no. 8Suppose that P [B | A1] = 1/2 and that P [B | A2] = 1/4, where A1 and A2 are two

equiprobable events forming a partition of Ω. Calculate P [A1 | B].

Question no. 9Three horses, a, b, and c, enter in a race. If the outcome bac means that b finished

first, a second, and c third, then the set of all possible outcomes is

Ω = abc, acb, bac, bca, cab, cba .

We suppose that P [abc] = P [acb] = 1/18 and that each of the other four elementaryoutcomes has a 2/9 probability of occurring. Moreover, we define the events

A = “a finishes before b” and B = “a finishes before c.”

(a) Do the events A and B form a partition of Ω?(b) Are A and B independent events?

Question no. 10Let ε be a random experiment for which there are three elementary outcomes: A, B,

and C. Suppose that we repeat ε indefinitely and independently. Calculate, in terms ofP [A] and P [B], the probability that A occurs before B.Hints. (i) You can make use of the law of total probability.(ii) Let D be the event “A occurs before B.” Then, we may write that

P [D | C occurs on the first repetition] = P [D] .

Question no. 11Transistors are drawn at random and with replacement from a box containing a

very large number of transistors, some of which are defectless and others are defective,and are tested one at a time. We continue until either a defective transistor has beenobtained or three transistors in all have been tested. Describe the sample space Ω forthis random experiment.

Question no. 12Let A and B be events such that P [A] = 1/3 and P [B′ | A] = 5/7. Calculate P [B]

if B is a subset of A.

Question no. 13In a certain university, the proportion of full, associate, and assistant professors,

and of lecturers is 30%, 40%, 20%, and 10% respectively, of which 60%, 70%, 90%, and40% hold a PhD. What is the probability that a person taken at random among thoseteaching at this university holds a PhD?

Question no. 14All the items in stock in a certain store bear a code made up of five letters. If the

same letter is never used more than once in a given code, how many different codes canthere be?


Question no. 15A fair die is rolled twice, independently. Consider the eventsA = “the first number that shows up is a 6;”B = “the sum of the two numbers obtained is equal to 7;”C = “the sum of the two numbers obtained is equal to 7 or 11.”

(a) Calculate P [B | C].(b) Calculate P [A | B].(c) Are A and B independent events?

Question no. 16A commuter has two vehicles, one being a compact car and the other one a minivan.

Three times out of four, he uses the compact car to go to work and the remainder ofthe time he uses the minivan. When he uses the compact car (resp., the minivan), hegets home before 5:30 p.m. 75% (resp., 60%) of the time. However, the minivan has airconditioning. Calculate the probability that(a) he gets home before 5:30 p.m. on a given day;(b) he used the compact car if he did not get home before 5:30 p.m.;(c) he uses the minivan and he gets home after 5:30 p.m.;(d) he gets home before 5:30 p.m. on two (independent) consecutive days and he doesnot use the same vehicle on these two days.

Question no. 17Rain is forecast half the time in a certain region during a given time period. We

estimate that the weather forecasts are accurate two times out of three. Mr. X goes outevery day and he really fears being caught in the rain without an umbrella. Consequently,he always carries his umbrella if rain is forecast. Moreover, he even carries his umbrellaone time out of three if rain is not forecast. Calculate the probability that it is rainingand Mr. X does not have his umbrella.

Question no. 18A fair die is rolled three times, independently. Let F be the event “the first number

obtained is smaller than the second one, which is itself smaller than the third one.”Calculate P [F ].

Question no. 19We consider the set of all families having exactly two children. We suppose that each

child has a 50–50 chance of being a boy. Let the events beA1 = “both sexes are represented among the children;”A2 = “at most one child is a girl.”

(a) Are A1 and A′2 incompatible events?

(b) Are A1 and A′2 independent events?

(c) We also suppose that the probability that the third child of an arbitrary family is aboy is equal to 11/20 if the first two children are boys, to 2/5 if the first two children


are girls, and to 1/2 in the other cases. Knowing that the third child of a given familyis a boy, what is the probability that the first two are also boys?

Exercises

Question no. 1We study the traffic (in one direction) on two roads, 1 and 2, which merge to form

road 3 (see Figure 2.12). Roads 1 and 2 have the same capacity (number of lanes) androad 3 has a greater capacity than road 1 and road 2. During rush hours, the probabilitythat the traffic is congested on road 1 (resp., road 2) is equal to 0.1 (resp., 0.3). Moreover,given that traffic is congested on road 2, it is also congested on road 1 one time out ofthree. We define the events

A,B, C = “traffic is congested on roads 1, 2, 3, respectively.”

1

23

Fig. 2.12. Figure for Exercise no. 1.

(a) Calculate the probability that traffic is congested(i) on roads 1 and 2;(ii) on road 2, given that it is congested on road 1;(iii) on road 1 only;(iv) on road 2 only;(v) on road 1 or on road 2;(vi) neither on road 1 nor on road 2.

(b) On road 3, traffic is congested with probability

1 if it is congested on roads 1 and 2;0.15 if it is congested on road 2 only;0.1 if it is neither congested on road 1 nor on road 2.

Calculate the probability that traffic is congested(i) on road 3;(ii) on road 1, given that it is congested on road 3.


Question no. 2We roll a die and then we toss a coin. If we obtain “tails,” then we roll the die a

second time. Suppose that the die and the coin are fair. What is the probability of(a) obtaining “heads” or a 6 on the first roll of the die;(b) obtaining no 6s;(c) obtaining exactly one 6;(d) having obtained “heads,” given that we obtained exactly one 6.

Question no. 3 (see Example 2.4.1)A device is composed of two components, A and B, subject to random failures. The

components are connected in parallel and, consequently, the device is down only if bothcomponents are down. The two components are not independent. We estimate that theprobability of

a failure of component A is equal to 0.2;a failure of component B is equal to 0.8 if component A is down;a failure of component B is equal to 0.4 if component A is active.

(a) Calculate the probability of a failure(i) of component A if component B is down;(ii) of exactly one component.

(b) In order to increase the reliability of the device, a third component, C, is added insuch a way that components A, B, and C are connected in parallel. The probabilitythat component C breaks down is equal to 0.2, independently of the state (up or down)of components A and B. Given that the device is active, what is the probability thatcomponent C is down?

Question no. 4In a factory producing electronic parts, the quality control is ensured through three

tests as follows:

• each component is subjected to test no. 1;• if a component passes test no. 1, then it is subjected to test no. 2;• if a component passes test no. 2, then it is subjected to test no. 3;• as soon as a component fails a test, it is returned for repair.

We define the events

Ai = “the component fails test no. i, for i = 1, 2, 3.”

From past experience, we estimate that

P [A1] = 0.1, P [A2 | A′1] = 0.05 and P [A3 | A′

1 ∩A′2] = 0.02.

The elementary outcomes of the sample space Ω are: ω1 = A1, ω2 = A′1 ∩ A2, ω3 =

A′1 ∩A′

2 ∩A3, and ω4 = A′1 ∩A′

2 ∩A′3.


(a) Calculate the probability of each elementary outcome.(b) Let R be the event “the component must be repaired.”

(i) Express R in terms of A1, A2, A3.(ii) Calculate the probability of R.(iii) Calculate P [A′

1 ∩A2 | R].

(c) We test three components and we define the events

Rk = “the kth component must be repaired, for k = 1, 2, 3” and

B = “at least one of the three components passes all three tests.”

We assume that the events Rk are independent.(i) Express B in terms of R1, R2, R3.(ii) Calculate P [B].

Question no. 5Let A, B, and C be events such that P [A] = 1/2, P [B] = 1/3, P [C] = 1/4, and

P [A ∩ C] = 1/12. Furthermore, A and B are incompatible. Calculate P [A | B ∪ C].

Question no. 6In a group of 20,000 men and 10,000 women, 6% of men and 3% of women suffer

from a certain disease. What is the probability that a member of this group sufferingfrom the disease in question is a man?

Question no. 7Two tokens are taken at random and without replacement from an urn containing

10 tokens, numbered from 1 to 10. What is the probability that the larger of the twonumbers obtained is 3?

Question no. 8We consider the system in Figure 2.13. All components fail independently of each

other. During a certain time period, the type A components fail with probability 0.3 andcomponent B (resp., C) fails with probability 0.01 (resp., 0.1). Calculate the probabilitythat the system is not down at the end of this period.

Question no. 9A sample of size 20 is drawn (without replacement) from a lot of infinite size con-

taining 2% defective items. Calculate the probability of obtaining at least one defectiveitem in the sample.

Question no. 10A lot contains 10 items, of which one is defective. The items are examined one

by one, without replacement, until the defective item has been found. What is theprobability that this defective item will be (a) the second item examined? (b) the ninthitem examined?


A

BA

A

C


Question no. 11A bag holds two coins: a fair one and one with which we always get “heads.” A coin

is drawn at random and is tossed. Knowing that “heads” was obtained, calculate(a) the probability that the fair coin was drawn;(b) the probability of obtaining “heads” on a second toss of the same coin.

Question no. 12The diagnosis of a physician in regard to one of her patients is unsure. She hesitates

between three possible diseases. From past experience, we were able to construct thefollowing tables:

Si S1 S2 S3 S4

P [D1 | Si] 0.2 0.1 0.6 0.4

Si S1 S2 S3 S4

P [D2 | Si] 0.2 0.5 0.5 0.3

Si S1 S2 S3 S4

P [D3 | Si] 0.6 0.3 0.1 0.2

where the Dis represent the diseases and the Sis are the symptoms. In addition, weassume that the four symptoms are incompatible, exhaustive, and equiprobable.(a) Independently of the symptom present in the patient, what is the probability thathe or she suffers from the first disease?(b) What is the probability that the patient suffers from the second disease and presentssymptom S1?(c) Given that the patient suffers from the third disease, what is the probability thathe or she presents symptom S2?(d) We consider two independent patients. What is the probability that they do notsuffer from the same disease, if we assume that the three diseases are incompatible?


Question no. 13We consider a system comprising four components operating independently of each

other and connected as in Figure 2.14.

4

21

3


The reliability of each component is supposed constant, over a certain time period,and is given by the following table:

Component 1 2 3 4Reliability 0.9 0.95 0.95 0.99

(a) What is the probability that the system operates at the end of this time period?(b) What is the probability that component no. 3 is down and the system still operates?(c) What is the probability that at least one of the four components is down?(d) Given that the system is down, what is the probability that it will resume operatingif we replace component no. 1 by an identical (nondefective) component?

Question no. 14A box contains 8 brand A and 12 brand B transistors. Two transistors are drawn

at random and without replacement. What is the probability that they are both of thesame brand?

Question no. 15What is the reliability of the system shown in Figure 2.15 if the four components

operate independently of each other and all have a reliability equal to 0.9 at a giventime instant?


4

2

1

3


Question no. 16Let A1 and A2 be two events such that P [A1] = 1/4, P [A1 ∩A2] = 3/16, and

P [A2 | A′1] = 1/8. Calculate P [A′

2].

Question no. 17A fair coin is tossed until either “heads” is obtained or the total number of tosses

is equal to 3. Given that the random experiment ended with “heads,” what is theprobability that the coin was tossed only once?

Question no. 18In a room, there are four 18-year-old male students, six 18-year-old female students,

six 19-year-old male students, and x 19-year-old female students. What must be thevalue of x, if we want age and sex to be independent when a student is taken at randomin the room?

Question no. 19Stores S1, S2, and S3 of the same company have, respectively, 50, 70, and 100

employees, of which 50%, 60%, and 75% are women. A person working for this companyis taken at random. If the employee selected is a woman, what is the probability thatshe works in store S3?

Question no. 20Harmful nitrogen oxides constitute 20%, in terms of weight, of all pollutants present

in the air in a certain metropolitan area. Emissions from car exhausts are responsiblefor 70% of these nitrogen oxides, but for only 10% of all the other air pollutants. Whatpercentage of the total pollution for which emissions from car exhausts are responsibleare harmful nitrogen oxides?

Question no. 21Three machines, M1, M2, and M3, produce, respectively, 1%, 3%, and 5% defective

items. Moreover, machine M1 produces twice as many items on an arbitrary day asmachine M2, which itself produces three times as many items as machine M3. An itemis taken at random among those manufactured on a given day, then a second item istaken at random among those manufactured by the machine that produced the first


selected item. Knowing that the first selected item is defective, what is the probabilitythat the second one is also defective?

Question no. 22A machine is made up of five components connected as in the diagram of Figure 2.16.

Each component operates with probability 0.9, independently of the other components.

4

2

1

3

5


(a) Knowing that component no. 1 is down, what is the probability that the machineoperates?(b) Knowing that component no. 1 is down and that the machine still operates, whatis the probability that component no. 3 is active?

Question no. 23Before being declared to conform to the technical norms, devices must pass two

quality control tests. According to the data gathered so far, 75% of the devices testedin the course of a given week passed the first test. The devices are subjected to thesecond test, whether they pass the first test or not. We found that 80% of the devicesthat passed the second test had also passed the first one. Furthermore, 20% of thosethat failed the second test had passed the first one.(a) What is the probability that a given device passed the second test?(b) Find the probability that, for a given device, the second test contradicts the firstone.(c) Calculate the probability that a given device failed the second test, knowing that itpassed the first one.

Question no. 24In a certain workshop, the probability that a part manufactured by an arbitrary

machine is nondefective, that is, conforms to the technical norms, is equal to 0.9. Thequality control engineer proposes to adopt a procedure that classifies as nondefectivewith probability 0.95 the parts indeed conforming to the norms, and with only prob-ability 0.15 those not conforming to these norms. It is decided that every part will besubjected to this quality control procedure twice, independently.


(a) What is the probability that a part having passed the procedure twice does indeedconform to the norms?(b) Suppose that if a part fails the first control test, then it is withdrawn immediately.Let Bj be the event “a given part passed (if the case may be) the jth control test,” forj = 1, 2. Calculate (i) P [B2] and (ii) P [B′

1 ∩B′2].

Question no. 25We have 20 type I components, of which 5 are defective, and 30 type II components,

of which 15 are defective.

(a) We wish to construct a system comprising 10 type I components and 5 type IIcomponents connected in series. What is the probability that the system will operate ifthe components are taken at random?(b) How many different systems comprising four components connected in series, ofwhich at least two are of type I, can be constructed, if the order of the components istaken into account?Remarks. (i) We suppose that we can differentiate two components of the same type.(ii) When a system is made up of components connected in series, then it operates ifand only if every component operates.

Question no. 26A system is made up of n components, including components A and B.

(a) Show that if the components are connected in series, then the probability that thereare exactly r components between A and B is given by

2(n− r − 1)(n− 1)n

for r ∈ 0, 1, . . . , n− 2.

(b) Calculate the probability that there are exactly r components between A and B ifthe components are connected in circle.(c) Suppose that n = 5 and that the components are connected in series. Calculatethe probability of operation of the subsystem constituted of components A, B and ther components placed between them if the components operate independently of eachother and all have a reliability of 0.95.

Question no. 27A man owns a car and a motorcycle. Half the time, he uses his motorcycle to go to

work. One-third of the time, he drives his car to work and, the remainder of the time,he uses public transportation. He gets home before 5:30 p.m. 75% of the time when heuses his motorcycle. This percentage is equal to 60% when he drives his car and to 2%when he uses public transportation. Calculate the probability that(a) he used public transportation if he got home after 5:30 p.m. on a given day;(b) he got home before 5:30 p.m. on two consecutive (independent) days and he usedpublic transportation on exactly one of these two days.


Question no. 28In a certain airport, a shuttle coming from the city center stops at each of the four

terminals to let passengers get off. Suppose that the probability that a given passengergets off at a particular terminal is equal to 1/4. If there are 20 passengers using theshuttle and if they occupy seats numbered from 1 to 20, what is the probability thatthe passengers sitting in seats nos. 1 to 4 all get off(a) at the same stop?(b) at different stops?

Question no. 29A square grid consists of 289 points. A particle is at the center of the grid. Every

second, it moves at random to one of the four nearest points from the one it occupies.When the particle arrives at the boundary of the grid, it is absorbed.(a) What is the probability that the particle is absorbed after eight seconds?(b) Let Ai be the event “the particle is at the center of the grid after i seconds.”Calculate P [A4] (knowing that A0 is certain, by assumption).

Question no. 30Five married couples bought 10 tickets for a concert. In how many ways can they

sit (in the same row) if(a) the five men want to sit together?(b) the two spouses in each couple want to sit together?

Multiple choice questions

Question no. 1Two weeks prior to the most recent general election, a poll conducted among 1000

voters revealed that 48% of them intended to vote for the party in power. A survey madeafter the election, among the same sample of voters, showed that 90% of the personswho indeed voted for the party in power intended to vote for this party, and 10% ofthose who voted for another party intended (two weeks prior to the election) to vote forthe party in power. Let the events be

A = “a voter, taken at random in the sample, intended to vote for the party inpower;”

B = “a voter, taken at random in the sample, voted for the party in power.”(A) From the statement of the problem, we can write that P [A] = 0.48 and that(a) P [A ∩B] = 0.9; P [A ∩B′] = 0.1(b) P [B | A] = 0.9; P [B′ | A] = 0.1(c) P [A | B] = 0.9; P [A | B′] = 0.1(d) P [A′ ∩B] = 0.9; P [A ∩B′] = 0.1(e) P [A | B] = 0.9; P [B′ | A] = 0.1


(B) The probability of event B is given by(a) 0.45 (b) 0.475 (c) 0.48 (d) 0.485 (e) 0.50 (f) 0.515(C) Are events A and B′ incompatible?(a) yes (b) no (c) we cannot conclude from the information provided(D) Are events A and B′ independent?(a) yes (b) no (c) we cannot conclude from the information provided(E) Do events A and B′ form a partition of the sample space Ω?(a) yes (b) no (c) we cannot conclude from the information provided(F) Let E be “a voter, taken at random among the 1000 voters polled, did not vote ashe intended to two weeks prior to the election (in regard to the party in power).” Wecan write that(a) P [E] = P [A | B′] + P [A′ | B](b) P [E] = P [A ∩B] + P [A′ ∩B′](c) P [E] = P [B′ | A] + P [B | A′](d) P [E] = P [A ∩B′] + P [A′ ∩B′](e) P [E] = P [A ∩B′] + P [A′ ∩B]

Question no. 2Let A and B be two events such that

P [A ∩B] = P [A′ ∩B] = P [A ∩B′] = p.

Calculate P [A ∪B].(a) p (b) 2p (c) 3p (d) 3p2 (e) p3

Question no. 3We have nine electronic components, of which one is defective. Five components are

taken at random to construct a system in series. What is the probability that the systemdoes not operate?(a) 1/3 (b) 4/9 (c) 1/2 (d) 5/9 (e) 2/3

Question no. 4Two dice are rolled simultaneously. If a sum of 7 or 11 is obtained, then a coin is

tossed. How many elementary outcomes [of the form (die1, die2) or (die1, die2, coin)]are there in the sample space Ω?(a) 28 (b) 30 (c) 36 (d) 44 (e) 72

Question no. 5Let A and B be two independent events such that P [A] < P [B], P [A ∩B] = 6/25,

and P [A | B] + P [B | A] = 1. Calculate P [A].(a) 1/25 (b) 1/5 (c) 6/25 (d) 2/5 (e) 3/5


Question no. 6In a certain lottery, 4 balls are drawn at random and without replacement among

25 balls numbered from 1 to 25. The player wins the grand prize if the 4 balls thatshe selected are drawn in the order indicated on her ticket. What is the probability ofwinning the grand prize?

(a)1

12, 650(b)

24390, 625

(c)1

303, 600(d)

1390, 625

(e)1

6, 375, 600

Question no. 7New license plates are made up of three letters followed by three digits. If we suppose

that the letters I and O are not used and that no plates bear the digits 000, how manydifferent plates can there be?(a) 243 × 93 (b) (26× 25× 24) (10× 9× 8) (c) 243 × (10× 9× 8)(d) 243 × 999 (e) 253 × 93

Question no. 8Let P [A | B] = 1/2, P [B′] = 1/3, and P [A ∩B′] = 1/4. Calculate P [A].

(a) 1/4 (b) 1/3 (c) 5/12 (d) 1/2 (e) 7/12

Question no. 9In the lottery known as 6/49, first 6 balls are drawn at random and without replace-

ment among 49 balls numbered from 1 to 49. Next, a seventh ball (the bonus number)is drawn at random among the 43 remaining balls. A woman selected what she thinkswould be the six winning numbers for the next draw. What is the probability that thiswoman actually did not select any of the seven balls that will be drawn (including thebonus number)?

(a)

(426

)(496

) (b)

(427

)(496

) (c)

(426

)(497

) (d)

(436

)(497

) (e)

(427

)(497

)Question no. 10

In a quality control procedure, every electronic component manufactured is subjectedto (at most) three tests. After the first test, an arbitrary component is classified as either“good,” “average,” or “defective,” and likewise after the second test. Finally, after thelast test, the components are classified as either “good” or “defective.” As soon as acomponent is classified as defective after a test, it is returned to the factory for repair.The following random experiment is performed: a component is taken at random andthe result of each test it is subjected to is recorded. How many elementary outcomesare there in the sample space Ω?(a) 3 (b) 8 (c) 11 (d) 18 (e) 21

Question no. 11Let P [A] = 1/3, P [B] = 1/2, P [C] = 1/4, P [A | B] = 1/2, P [B | A] = 3/4,

P [A | C] = 1/3, P [C | A] = 1/4, and P [B ∩ C] = 0. Calculate the probabilityP [A | B ∪ C].


(a) 0 (b) 1/3 (c) 4/9 (d) 5/6 (e) 1

Question no. 12A fair die is rolled twice (independently). Consider the eventsA = “the two numbers obtained are different;”B = “the first number obtained is a 6;”C = “the two numbers obtained are even.”

Which pairs of events are the only ones comprised of independent events?(a) no pairs (b) (A,B) (c) (A,B) and (B,C) (d) (A,B) and (A,C)(e) the three pairs

Question no. 13A man plays a series of games for which the probability of winning a given game,

from the second one, is equal to 3/4 if he won the previous game and to 1/4 otherwise.Calculate the probability that he wins games nos. 2 and 3 consecutively if the probabilitythat he wins the first game is equal to 1/2.(a) 3/16 (b) 1/4 (c) 3/8 (d) 9/16 (e) 5/8

Question no. 14A box contains two coins, one of them being fair but the other one having two

“heads.” A coin is taken at random and is tossed twice, independently. Calculate theprobability that the fair coin was selected if “heads” was obtained twice.(a) 1/5 (b) 1/4 (c) 1/3 (d) 1/2 (e) 3/5

http://www.springer.com/978-0-387-74994-5

Documents

2 Elementary probability